The number of real solutions of the equation x2+4x+3+2x+5=0 are

We have,x2+4x+3+2x+5=0Here, two cases arise.Case l  When x2 +4x+3>0x2+4x+3+2x+5=0x2+6x+8=0(x+2)(x+4)=0x=−2,−4x = -2 is not satisfying the condition x2 + 4x + 3 > 0.So, x= - 4 is the only solution of the given equation. Case ll When x2+4x+3<0⇒        −x2+4x+3+2x+5=0⇒    −x2−2x+2=0⇒        x2+2x−2=0⇒        (x+1+3)(x+1−3)=0⇒        x=−1+3,−1−3Hence, x=−(1+3) satisfy the given condition. Since,x2+4x+3<0 whilex=−1+3 is not satisfying the condition. Thus, number of real solutions are two.